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\title{\huge Clothoid \footnote{This file is from the 3D-XploreMath project. \hfil\break Please see http://rsp.math.brandeis.edu/3D-XplorMath/index.html}}
\author{}
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\maketitle

\LARGE


\vskip -35pt
\centerline{\includegraphics{clothoid3dxm.png}}

The Clothoid, also called Spiral of Cornu, is a curve whose curvature is equal to its arclength. 
It has the parametric formula:

$$ \left(  \int_0^t \cos(x^2/2)\,dx,
      \int_0^t \sin(x^2/2)\, dx) \right ) .$$

\goodbreak
  \centerline{\bf Discussion}
  
If a plane curve is given by a parametric 
formula $(f(t),g(t))$, then the length of the part corresponding to a parameter  
interval  $[a,t]$ is $ s(t) = \int_a^t \sqrt{ f'(\tau)^2 + g'(\tau)^2} \,d\tau$. If we 
apply this formula to the Clothoid we see that the arclength corresponding to
the interval $[0,t]$ is   $s(t) =  \int_0^t 1 \, dt = t$, so that the parameter $t$ 
is precisely the (signed!) arclength measured along the curve from its 
midpoint, $(0,0)$.


Next, recall that the curvature $\kappa$ of a plane curve is defined as the rate of change 
(with respect to arclength) of the angle $\theta$ that its tangent makes with some 
fixed line (which we can take to be the $x$-axis). And since the slope $dy \over dx$ of the curve is $\tan(\theta)$, and by the chain rule 
${dy \over dx} = (dy/dt)/(dx/dt) = {g'\over f'}$, 
we see that $\theta(t) = \arctan(g'(t)/f'(t))$. So if we assume that parameter $t$ is arclength, then using the formulas for the derivative of the arctangent and 
of a quotient, we see that:
$$
 \kappa(t) = \theta'(t) =   -g'(t)  f''(t) + f'(t) g''(t) ,
$$
(where we have ignored the denominator, since parameterization by arclength
implies that it equals unity).
Applying this to the Clothoid, we obtain $\kappa(t) = t$. Since the arclength function is also t, this shows that the Clothoid is indeed a curve whose curvature function  is equal to its 
arclength function.

\medskip
 \centerline{\bf The Fundamental Theorem of Plane Curves}

  Next let's  look at this question from the other direction, and also more generally.
  Suppose we are given a function $\kappa(t)$. Can we find a plane curve 
  parameterized by arclength $(f(t),g(t))$ such that $\kappa$ is its 
  curvature function? Recall from above that ${d\theta\over dt} = \kappa$,
  and of course ${dx\over dt} = f'(t)$ and ${dy\over dt} = g'(t)$. 
  Now, since $({dx\over dt})^2 +  ({dy\over dt})^2 = 1$, while 
  $ {dy\over dt}/{dx\over dt} = dy/dx = \tan(\theta)$, it follows from 
  elementary trigonometry that ${dx\over dt} = \cos(\theta)$ while
 ${dy\over dt} = \sin(\theta)$. Thus we have the following system of three 
 differential equations for the three functions $\theta(t)$, $f(t)$, and $g(t)$:
  \begin{align*}
   \theta'(t) &= \kappa(t)    \\
      f'(t) &= \cos(\theta(t))  \\
      g'(t) &= \sin(\theta(t)).   \\
  \end{align*}
  The first equation has the general solution 
  $\theta(\tau) = \theta_0 + \int_0^\tau \kappa(\sigma) \, d\sigma$,
  and substituting this in the other two equations, we find that the 
  general solutions for $f$ and $g$ are given by:
\begin{align*}
   f(t) &= x_0 + \int_0^t \cos(\theta_0 + \int_0^\tau \kappa(\sigma) \, d\sigma)\,d\tau  \\
   g(t) &= y_0 + \int_0^t \sin(\theta_0 + \int_0^\tau \kappa(\sigma) \, d\sigma)\,d\tau.   \\
  \end{align*}
This is an elegant explicit solution to our question! It shows that not only is there a 
solution to our question (say the one obtained by setting $x_0,y_0$ and $\theta_0$ all 
equal to zero), but also that the solution is unique up to a translation (by $(x_0,y_0)$)
and a rotation (by $\theta_0$), that is unique up to a general rigid motion.

  This fact has a name---it is called The Fundamental Theorem of Plane Curves. 
It tell us us that most geometric and most economical descriptions of plane curves
is not via parametric equations, which have a lot of redundancy, but rather by the
single function $\kappa$ that gives the curvature as a function of arclength.
  
{\bf Exercise} Take $\kappa(t) = t$ and check that the above formulas give the parametric  equations for the Clothoid in this case.

\vfil
\eject
\centerline{\bf Back to the Clothoid}
\smallskip
We close with a few more details about the Clothoid. First, here is a plot of
the integrand $\sin(x^2/2)$: 
\bigskip

\centerline{\includegraphics{clothoid_integrand.png}}
\bigskip

and next a plot of its indefinite integral, $\int_0^t \sin(x^2/2) \,dx$,  the so-called Fresnel integral:
\bigskip

\centerline{\includegraphics{clothoid_fresnelS.png}}

 From this plot we see that the y-coordinate oscillates. Its limit as t goes to infinity is 
 $\sqrt{\pi}/2$, from which we see that the centers of the two spirals of the Clothoid 
 are at $\pm(\sqrt{\pi}/2,\sqrt{\pi}/2)$.

XL \& RSP.


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